limit $f(x)=((x-7)(x+4))^{1/2}$ I have to calculate the Asymptotes in the infinity(and minus infinity) for this function:
$f(x)=((x-7)(x+4))^{1/2}$
I know that
$\lim_{x \to\infty} f(x)/x= 1$
And I get into trouble with:
$\lim_{x \to\infty} f(x)-x$
which is
$\lim_{x \to\infty} ((x-7)(x+4))^{1/2}-x = \lim_{x \to\infty} \sqrt{x^2-3x-28}-x$
wolfram alpha says it is $-3/2$ but i don't get why.... please help me with that, thanks
 A: $$
\sqrt{x^2-3x-28} - x = \frac{(\sqrt{x^2-3x-28} - x)(\sqrt{x^2-3x-28} + x)}{(\sqrt{x^2-3x-28} + x)}\\
= \frac{x^2-3x-28-x^2}{(\sqrt{x^2-3x-28} + x)}
= \frac{-3x-28}{x(\sqrt{1-3/x-28/x^2} + 1)} \to \frac {-3}{2}
$$
A: The limit $$\lim_{x\to\infty} ((x−7)(x+4))^{1/2}$$ does not go to one. We can see this in a couple of ways. Let's take out $x^2$ from the radical, then we get: $$\lim_{x\to\infty} |x| \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)}$$
Then as $x \to \infty$ we see that the terms inside of the radical go to 1, and the $|x|$ goes to $+\infty$. Thus the limit diverges to $\infty$.
The same happens as $x \to -\infty$.
If you are trying to find the oblique asymptote, that is the long run behavior of $f$, then from what we have said so far, it looks like $f$ is tending to $x$ as $x\to\infty$. We can demonstrate this by writing:
$$\lim_{x\to\infty} \left(|x| \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - x\right)$$
$$=\lim_{x\to\infty} x\left( \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - 1\right)$$
$$=\lim_{x\to\infty} x\left( \frac{ \left(1-\frac7x\right)\left(1+\frac4x\right) - 1}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$
$$=\lim_{x \to\infty} \frac1x\left( \frac{ \left(x-7\right)\left(x+4\right) - x^2}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$
$$=\lim_{x \to\infty} \frac1x\left( \frac{ x^2 -3x - 28 - x^2}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$
$$=\lim_{x \to\infty} \left( \frac{  -3 - \frac{28}x}{\sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} + 1}\right)$$
$$=-3/2$$
A: Let me start from what Joel answered. We are concerned by the behavior of $$ x\left( \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - 1\right)$$ for large values of $x$. Let us rewrite it as 
$$x\left(\sqrt{ 1-\frac7x} \sqrt{ 1+\frac4x}-1\right)$$ and now use $$\sqrt{ 1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Apply it for each radical and replace $y$ by $-\frac{7}{x}$ in the first and by $\frac{4}{x}$ in the second; now develop to get $$x\left(1-\frac{3}{2 x}-\frac{121}{8 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)-1\right)$$ and finally obtain $$ x\left( \sqrt{ \left(1-\frac7x\right)\left(1+\frac4x\right)} - 1\right)=-\frac{3}{2}-\frac{121}{8 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$ So, you have the asymptote and moreover how the curve is with respect to it.
A: $$\lim_{x \to\infty} \sqrt{x^2-3x-28}-x=\lim_{x \to\infty} x\Big(\sqrt{1-(\frac3x+\frac{28}{x^2})}-1\Big).$$
It is enough to limit the development of $\sqrt{1-t}=1-\frac 12t+O(t^2)$ to the term in $\frac1x$.
$$\lim_{x \to\infty} x\Big(1-\frac3{2x}+O(\frac1{x^2})-1\Big)=-\frac32.$$
